I told you about step 2 and step 4. But, I should have been more descriptive [since you are kinda new to php].
But, in the end it doesn't even matter :P Congrats on finding the solution!
Cheers,...

Indeed :P

Cheers man :)

If it is *really* something small, then maybe its time for me to get a new pair of eyes. :D
Anyways, Lets hope someone else here finds what's wrong.
Anyways, Good luck :).

That's really strange. :S Sorry, It seems like I have reached the dead end and can't think of one good reason why it's failing to update the record.

Hmm.. I see what you are talking about. Did you try printing the query and executing it in phpmyadmin or mysql console ?
Your script looks good though :S

Well, I checked your script and it is fine. I don't see any errors. Apart from saying 'It works', does it say anything else ? Could you be more specific ?

Pass $row['date'] as a parameter to this date function and return the changed value.
Eg.

function changeDate($date) {
$newdate = date('d/n/y',strtotime($date));
return $newdate;
}

Call...

@evstevemd,
You missed "/" in myfoldername variable. So it will create the sql file as,

@futhonguy,
You can use evstevemd's code.

$myfoldername = getcwd()."\\MyBackups/";//your folders...

You have to mention the path where you want the .sql file to be created.
This will create a folder called backup [if it doesn't exist and then create the .sql file in that folder].

<?php...

:) As the error itself says, foto is undefined.

$ficheiro=$_FILES['foto'];

should be

$ficheiro=$_FILES['file'];

One more thing, notices are usually generated when you don't initialize a...

Hey, Hi.. Instead of ereg_replace, you an use str_replace. Both serve the same purpose. Replace \n with \\n so it don't get escaped. Anyways, the second 'error' is not an error but a notice. You can...

Ah, nevermind. My apologies.

Great :) Cheers!

Are you sure its administrator ? I know about mysql's 'root', but never heard of administrator.
And when you say phpmydirectory, are you talking about phpmyadmin or something else ?
Also check...

This is wrong. Try,

$query = "SELECT * FROM Employee WHERE $method = '$search'";
$result = mysql_query($query);

Ah! All's well. :) Cheers!

Use

array_map('stripslashes,$_POST);
array_map('mysql_real_escape_string',$_POST);

Here is a quote from php.net (http://php.net/manual/en/language.references.pass.php).

The only problem with this is , if a user loses his interest midway [ie., after submitting 1st form], there will be a obsolete record in the table. Or else, this is good.

You can do it in 2 simple ways (maybe there are even more, but I am not aware of it).
You can either save the data of the first form in session after it is posted, and then, when the second form is...

What I meant was ,

function procRestSearch($searchQuery){
//Don't forget to have $searchquery defined before calling this function
$intSearchCounter = 0; //Used to set the arrays...

Initialize the array outside the while loop. The array is getting over-written everytime it enters the loop. So you are getting only the last record.
Secondly, notices can be ignored. If you use...

When the user submits the page, build your query depending upon the user's selection.
If the user has selected from_year and to_year, include that in your condition and so on.

Please be specific with your question.

The echo statement is incorrect.

echo "<a href=\"inventorid.php?inventorid=".$data2['inventorid']."\">".$data2['inventorid']." ".$data2['firstname']." ".$data2['lastname']."</a>\n";

is the...

Technically, it doesn't matter if you have only 1 line statement after a condition. ie.,

<?php
$x = 1;
if($x == 1)
echo "$x is valid.";
else
echo "$x is invalid";
?>

I didn't go through so many functions in your script, but it works. You are supposed to select the options from the 2nd list before submitting. Unless you select them, they are not selected by...

Not only echo, you will get this warning if you have session_start along with header function. Use ob_start(); as the first line of your script to turn on output buffering and ob_end_flush(); as...

Very simple.

<html>
<head>
<script type="text/javascript">
function checkCheckbox() {
var countselection = 0;
var maximumselection = 2;
var ch = document.getElementsByName('choice');
...

To get a new line, you should add a <br /> to your echo.
ie.,

<?php
echo $abbr['US']['MI']."<br />";
?>
//or simply
<?php
echo $abbr['US']['MI'];
echo "<br />";

ob_start, as it indicates, will start output buffering. If you are using ob_start, then you should also use ob_end_flush, ie., flushing the output buffer.
I have used ob_start at places when I had...

Cheers! :)

I am glad I could help :)

:) You forgot the column name as the 2nd parameter for mysql_result.
http://php.net/manual/en/function.mysql-result.php

Condition is a mysql reserved keyword. In order to use it, you have to use ` [don't confuse it for single quote]. ie., `condition`. In my opinion, use some other word instead of condition to avoid...

Yeah, ofcourse. :) $string is just a variable name. You can give any name to a variable.
See Php naming convention here (http://www.tizag.com/phpT/variable.php).

Clear @ before mysql_query and see if it throws any error. Also, you can have $result = mysql_query($query) or die(mysql_error()); to print the error when the query fails.

P.S. @ suppresses the...

Or simply, $string = implode(",",$_POST['Prefered_Engineers']);

Hmm.. You are right! That makes sense..

@cwarn23, I think the OP was saying, that it shows only 1 record instead of 2 !
@levsha,
Are you sure you have 2 records for that query ? If you have 2 records, it *should* display them. There is...